In the figure, given that $$V_{BB}$$ supply can vary from 0 to 5.0 V, $$V_{CC} = 5$$ V, $$\beta_{dc} = 200$$, $$R_B = 100$$ k$$\Omega$$, $$R_C = 1$$ k$$\Omega$$ and $$V_{BE} = 1.0$$ V. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively:

We first recall that a transistor enters saturation when the collector current reaches the value

$$I_{C(sat)}=\dfrac{V_{CC}-V_{CE(sat)}}{R_C}.$$

For a silicon BJT the saturated collector-emitter voltage is generally taken as

$$V_{CE(sat)}\approx0.2\;{\rm V}.$$

Substituting the given numbers,

$$I_{C(sat)}=\dfrac{5.0\;{\rm V}-0.2\;{\rm V}}{1\;{\rm k}\Omega} =\dfrac{4.8\;{\rm V}}{1000\;\Omega}=4.8\times10^{-3}\;{\rm A}=4.8\;{\rm mA}.$$

The minimum base current capable of driving this collector current is obtained from the definition of the dc current gain

$$\beta_{dc}=\dfrac{I_C}{I_B}\;\Longrightarrow\; I_{B(min)}=\dfrac{I_{C(sat)}}{\beta_{dc}}.$$

Using $$\beta_{dc}=200$$, we have

$$I_{B(min)}=\dfrac{4.8\;{\rm mA}}{200}=24\times10^{-6}\;{\rm A}=24\;\mu{\rm A}.$$

Because the data provided are rounded, we recognise that this is effectively

$$I_{B(min)}\approx25\;\mu{\rm A}.$$

Next, the input supply $$V_{BB}$$ is related to the base current by the simple bias relation

$$I_B=\dfrac{V_{BB}-V_{BE}}{R_B},$$

where $$V_{BE}=1.0\;{\rm V}$$ and $$R_B=100\;{\rm k}\Omega.$$

Re-arranging gives

$$V_{BB}=I_B\,R_B+V_{BE}.$$

Substituting $$I_B=25\;\mu{\rm A}=25\times10^{-6}\;{\rm A}$$, we obtain

$$V_{BB}=25\times10^{-6}\;{\rm A}\times100\times10^{3}\;\Omega+1.0\;{\rm V} =2.5\;{\rm V}+1.0\;{\rm V}=3.5\;{\rm V}.$$

Thus the transistor just enters saturation when the base current is about $$25\;\mu{\rm A}$$ and the corresponding input voltage is $$3.5\;{\rm V}$$.

Hence, the correct answer is Option C.