Another answer says:

>In other words its inconclusive to determine if the transistor is either in the active, saturation or cut-off region. Since there is no reference made with the base terminal.

However, the base being open does not prevent us from calculating \$V_{BE}\$.

If the base is disconnected, then \$I_{B}=0\$. One can then use the Ebers-Moll equations, (or more sophisticated models) to find \$V_{BE}\$ from \$V_{CE}\$

We know that

$$V_{BC}=V_{BE}-V_{CE}$$

From the Ebers-Moll model,

$$I_{B} = I_{S}[\frac{1}{\beta_F}(e^{V_{BE}/V_T}-1) +\frac{1}{\beta_R}(e^{V_{BC}/V_T}-1)]$$

Setting \$I_B=0\$ and rearranging, gives

$$\beta_R(e^{V_{BE}/V_T} -1) + \beta_F(e^{V_{BC}/V_T} -1) = 0 $$

$$\beta_Re^{V_{BE}/V_T}  + \beta_Fe^{V_{BC}/V_T} = \beta_R + \beta_F $$

$$\beta_Re^{V_{BE}/V_T}  + \beta_Fe^{V_{BE}/V_T-V_{CE}/V_T} = \beta_R + \beta_F $$

$$e^{V_{BE}/V_T}[\beta_R  + \beta_Fe^{-V_{CE}/V_T}] = \beta_R + \beta_F $$

So, if I have done my math correctly,

$$e^{V_{BE}/V_T}=\frac{\beta_R + \beta_F} {\beta_R  + \beta_Fe^{-V_{CE}/V_T}} $$

If \$V_{CE}\$ is "large" relative to \$V_T\$, and \$\beta_F\$ is large relative to \$\beta_R\$ then the above approximates to

$$e^{V_{BE}/V_T} \approx \frac{\beta_F}{\beta_R}$$

or

$$V_{BE} \approx V_T \cdot ln(\frac{\beta_F}{\beta_R})$$

Choosing a random value of \$\frac{\beta_F}{\beta_R}\$ of 30, gives \$V_{BE}\approx 85\$ mV. Consistent with our intuition, when the base is open-circuit, the transistor is in the cutoff region. \$V_{BE}\$ is too small for the transistor to be in the forward active region. There will be some leakage current through the emitter and collector, but it will be relatively small.