Here, $A=500\, cm^{2} = 500 \times 10^{-4}\, m^{2}$
$N=1000, B=0.4\,G=0.4 \times 10^{-4}\,T$
$t=\frac{1}{10}\,s$
When the coil is held perpendicular to the field, the normal to the plane of the coil makes an angle of $0^{\circ}$ with the field $B$.
$\therefore$ initial flux, $\phi_{1}=BA\, cos\, 0^{\circ}=BA$
final flux, $\phi_{2}=BA\, cos\, 180^{\circ}=-BA$
$\varepsilon =-N (\frac{(\phi_{2}-\phi_{1})}{t})=-N (\frac{-BA-BA}{t})$
$=\frac{2NBA}{t}$
$=\frac{ 2\times1000\times0.4\times10^-4\times500\times10^{-4} }{ 1/10}$
$=0.04 \, V$