The voltages and currents in the circuit do not oscillate at a single frequency $\omega$. Your Fourier transform table is telling you that $v_g(t)$ in this case can be written as the superposition
$$ v_g(t) = \frac{1}{2\pi}\int\limits_{-\infty}^\infty V_g(\omega) e^{j\omega t}\, d\omega = \frac{1}{2\pi}\int\limits_{-\infty}^\infty \frac{36\times 2}{j\omega} e^{j\omega t}\, d\omega.$$
Recalling that voltage obeys the superposition principle, do you see what to do from here?
Expansion of the above: Forget for the moment that you've been asked to compute the response to \$v_g(t)\$; instead, suppose that the driving voltage is \$\tilde{v}_g(t) = A_g e^{j\omega t}\$. This is what you mentioned in your original question, although here I've switched to using complex phasor notation. Keep in mind that \$A_g\$ may be complex: \$A_g = |A_g|e^{j\phi_g}\$.
Using the Kirchoff loop laws and what you know about the impedances of resistors and capacitors, you should be able to find the sinusoidal voltage across the 60 kΩ resistor:
$$\tilde{v}_0(t) = A_0 e^{-j\omega t},$$
where \$A_0\$ is another complex number. You'll find that
$$ A_0 = f(\omega) A_g $$
for some function \$f(\omega)\$ [hint: it will be a rational function of \$\omega\$].
What does this mean? Well, your Fourier table has told you that the driving voltage \$v_g(t) = 36\mathrm{sgn}(t)\$ has an oscillatory component \$A_g e^{j\omega t}\$ with \$A_g = V_g(\omega) = 36\times 2/j\omega\$. Therefore, the voltage \$v_0(t)\$ has an oscillatory component \$A_0 e^{j\omega t}\$ with \$A_0 = V_0(\omega) = f(\omega) V_g(\omega)\$. Therefore, \$v_0(t)\$ must be given by
$$ v_0(t) = \frac{1}{2\pi}\int\limits_{-\infty}^\infty V_0(\omega) e^{j\omega t}\, d\omega = \frac{1}{2\pi}\int\limits_{-\infty}^\infty f(\omega) V_g(\omega) e^{j\omega t}\, d\omega.$$
Depending on your professor's/grader's expectations, you may be expected to find an analytic expression for this integral.
