simulate this circuit – Schematic created using CircuitLab
So, for series circuit to find resonance we have to have values of impedances (L and C) to be equal. And then resonant frequency can be calculated wl and 1/wc but this is combination of elements how to find resonant frequency here. Is there any formula for this? Zeq={(1*-jXc)/1-jXc} + jXl
Well, notice that the input impedance of your circuit is given by:
\begin{equation} \begin{split} \underline{\text{Z}}_{\space\text{i}}\left(\omega\right)&=\underline{\text{Z}}_{\space\text{L}}+\underline{\text{Z}}_{\space\text{C}\space\text{||}\space\text{R}}\\ \\ &=\text{j}\omega\text{L}+\left(\frac{1}{\text{j}\omega\text{C}}\space\text{||}\space\text{R}\right)\\ \\ &=\text{j}\omega\text{L}+\frac{\displaystyle\frac{1}{\text{j}\omega\text{C}}\cdot\text{R}}{\displaystyle\frac{1}{\text{j}\omega\text{C}}+\text{R}}\\ \\ &=\text{j}\omega\text{L}+\frac{\displaystyle\frac{\text{j}\omega\text{C}}{\text{j}\omega\text{C}}\cdot\text{R}}{\displaystyle\frac{\text{j}\omega\text{C}}{\text{j}\omega\text{C}}+\text{j}\omega\text{C}\text{R}}\\ \\ &=\text{L}\omega\text{j}+\frac{\displaystyle\text{R}}{\displaystyle1+\text{CR}\omega\text{j}}\\ \\ &=\text{L}\omega\text{j}+\frac{\displaystyle\text{R}}{\displaystyle1+\text{CR}\omega\text{j}}\cdot\frac{\displaystyle1-\text{CR}\omega\text{j}}{\displaystyle1-\text{CR}\omega\text{j}}\\ \\ &=\text{L}\omega\text{j}+\frac{\displaystyle\text{R}\left(1-\text{CR}\omega\text{j}\right)}{\displaystyle1^2+\left(\text{CR}\omega\right)^2}\\ \\ &=\text{L}\omega\text{j}+\frac{\displaystyle\text{R}-\text{CR}^2\omega\text{j}}{\displaystyle1+\left(\text{CR}\omega\right)^2}\\ \\ &=\text{L}\omega\text{j}+\frac{\displaystyle\text{R}}{\displaystyle1+\left(\text{CR}\omega\right)^2}-\frac{\displaystyle\text{CR}^2\omega}{\displaystyle1+\left(\text{CR}\omega\right)^2}\cdot\text{j}\\ \\ &=\frac{\displaystyle\text{R}}{\displaystyle1+\left(\text{CR}\omega\right)^2}+\left(\text{L}\omega-\frac{\displaystyle\text{CR}^2\omega}{\displaystyle1+\left(\text{CR}\omega\right)^2}\right)\text{j}\\ \\ &=\frac{\displaystyle\text{R}}{\displaystyle1+\left(\text{CR}\omega\right)^2}+\omega\left(\text{L}-\frac{\displaystyle\text{CR}^2}{\displaystyle1+\left(\text{CR}\omega\right)^2}\right)\text{j} \end{split}\tag1 \end{equation}
Where \$\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.
So, we can see that the amplitude of the input impedance is given by:
\begin{equation} \begin{split} \left|\underline{\text{Z}}_{\space\text{i}}\left(\omega\right)\right|&=\left|\frac{\displaystyle\text{R}}{\displaystyle1+\left(\text{CR}\omega\right)^2}+\omega\left(\text{L}-\frac{\displaystyle\text{CR}^2}{\displaystyle1+\left(\text{CR}\omega\right)^2}\right)\text{j}\right|\\ \\ &=\sqrt{\left(\frac{\displaystyle\text{R}}{\displaystyle1+\left(\text{CR}\omega\right)^2}\right)^2+\left(\omega\left(\text{L}-\frac{\displaystyle\text{CR}^2}{\displaystyle1+\left(\text{CR}\omega\right)^2}\right)\right)^2} \end{split}\tag2 \end{equation}
Solving:
$$\frac{\displaystyle\partial\left|\underline{\text{Z}}_{\space\text{i}}\left(\omega\right)\right|}{\displaystyle\partial\omega}=0\space\Longrightarrow\space\omega_0=\dots\tag3$$
Gives:
$$\omega_0=\frac{\displaystyle1}{\displaystyle\text{C}^2\text{R}}\cdot\sqrt{\frac{\text{R}}{\text{L}}\cdot\sqrt{\text{C}^5\left(2\text{L}+\text{CR}^2\right)}-\text{C}^2}\tag4$$
Using your values, we find:
$$\omega_0=\sqrt{2\sqrt{30}-1}\approx3.15507\space\text{rad/sec}\tag5$$
And:
$$\left|\underline{\text{Z}}_{\space\text{i}}\left(\omega_0\right)\right|=\frac{\sqrt{4\sqrt{30}-21}}{10}\approx0.0953364\space\Omega\tag6$$
