Answer
$I_s=3.2mA$
Work Step by Step
We know that
$I_p=\frac{V_s}{V_p}I_s$..........eq(1)
We also know that
$I_s=\frac{V_s}{R_s}=\frac{2.4}{15}=0.16$
We plug in the known values in eq(1) to obtain:
$I_s=\frac{2.4}{120}(0.16)=3.2\times 10^{-3}=3.2mA$
