The electric field $E$ at a distance $r$ from the centre of a charged sphere (where $r > R$) is given by $E = \frac{k|q|}{r^2}$. Given: $E = 1.5 \times 10^3$ N/C $r = 20$ cm $= 0.2$ m $k = 9 \times 10^9$ N m$^2$/C$^2$

Solving for $|q|$: $|q| = \frac{E r^2}{k} = \frac{(1.5 \times 10^3) (0.2)^2}{9 \times 10^9} = \frac{1.5 \times 0.04 \times 10^3}{9 \times 10^9} = \frac{0.06 \times 10^{-6}}{9} \times 10^9$ (wait, arithmetic check: $0.06 \times 10^3 / 9 \times 10^9 = 6.67 \times 10^{-9}$) $|q| \approx 6.67 \times 10^{-9}$ C $= 6.67$ nC. Since the electric field points radially inward, the charge must be negative. Thus, $q = -6.67$ nC. (See NCERT Physics Class 12, Exercise 1.20).