Given :
$l = 30 cm = 30 \times 10^{-2} m$
$A = 2.5 cm^2 = 25 \times 10^{-4} m^2$
$N = 500$
$I_1 = 2.5$
$I_2 = 0$
$dt = 10^{-3} s$
Induced emf $e = \frac{d \phi}{dt} = \frac{d}{dt} (BA)$
Magnetic field induction B at a point inside the long solenoid carrying current $I$ is
$B = \mu_o n I \; \; \; \; \; \; \; (n = \frac{N}{l})$
$e = N A. \frac{dB}{dt} = A . \frac{d}{dt} (\mu_o \frac{N}{l} I). \frac{dI}{dt}$
$e = 500 \times 25 \times 10^{-4} \times 4 \times 3.14 \times 10^{-7} \times \frac{500}{30 \times 10^{-2}} \times \frac{2.5}{10^{-3}}$
$\therefore e = 6.5 V$