Urea’s molar mass (NH2CONH2) = $2(1 \times 14+2 \times 1)+1 \times 12+1 \times 16=60 \mathrm{~g} \mathrm{~mol}^{-1}$
0.25 urea in a molar aqueous solution denotes:
$1000 \mathrm{~g}$ of water contains $0.25 \mathrm{~mol}=(0.25 \times 60) \mathrm{g}$ of urea $=15 \mathrm{~g}$ of urea i.e.,
( $1000+15) \mathrm{g}$ of solution contains $15 \mathrm{~g}$ of urea
Hence, $2.5 \mathrm{~kg}(2500 \mathrm{~g})$ of solution contains $=\frac{15 \times 2500}{1000+15} g$
$=36.95~\text{g}$
$=37~\text{g of urea (approx}\text{.) }$ Hence, total Urea needed is $37 \mathrm{~g}$.