$\frac{\text{di}}{\text{dt}}=10^3 \text{A}/\text{s}$



Therefore induced emf across inductance, $\left|\mathrm{e}\right|\text{=L}\frac{\text{di}}{\text{dt}}$
$\left|\mathrm{e}\right|=\left(5\times {10}^{-3}\right)\left({10}^3\right)=5 \mathrm{V}$

Since, the current is decreasing, the polarity of this emf would be so as to increase the existing current. The circuit can be redrawn as



Now, ${\mathrm{V}}_{\mathrm{A}}-5+15+5={\mathrm{V}}_{\mathrm{B}}$

 ${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=15 \mathrm{V}$