In the given circuit, the voltage across load resistance $\left(R_L\right)$ is:

Both diodes are in forward bias. In forward bias, potential drop across $Ge$ and $Si$ will be $0.3 \mathrm{V} \& 0.7 \mathrm{V}$ respectively.

Therefore, current in the circuit $i=\frac{15-0.3-0.7}{4\times 1000}=3.5 \mathrm{mA}$

Required potential drop across $R_L$, 
$V_{\mathrm{L}}=i{R}_{\mathrm{L}}=\left(3.5\times {10}^{-3}\right)\times \left(2.5\times 1000\right) \mathrm{V}$

$=8.75 \mathrm{V}$