a
When resistance of voltmeter is $\infty$
$\mathrm{V}_{1}=12 \times \frac{400}{1000}=4.8 \mathrm{\,V}$
When resistance of voltmeter taken into account
$\mathrm{V}_{2}=\frac{12 \times 300}{900}=4 \mathrm{\,V}$
error $=\frac{4.8-4}{4.8} \times 100 \%=\frac{0.8}{4.8} \times 100 \%=\frac{50}{3} \%=16.67 \%$