Explanation

(c) The efficiency of transformer
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{Energy\ obtained\ from\ the\ secondary\ coil}{Energy\ given\ to\ the\ primary\ coil}$
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta = \frac{Output\ power\ }{input\ power}\ $
or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \eta = \frac{V_{s}I_{s}}{V_{p}I_{p}}$
Given, $V_{s}I_{s} = 100\ W,\ V_{p} = V,I_{p} = 0.5\ A$
Hence, $\eta = \frac{100}{220 \times 0.5} = 0.90 = 90\%$