According to the average form of Newton's law of cooling: $\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$ where $T_s$ is the surrounding (room) temperature.

For the first $10\text{ minutes}$: $\frac{3T - 2T}{10} = K \left( \frac{3T + 2T}{2} - T \right)$ $\frac{T}{10} = K \left( \frac{5T}{2} - T \right)$ $\frac{T}{10} = K \left( \frac{3T}{2} \right)$ ... (i)

For the next $10\text{ minutes}$, let the final temperature be $T'$: $\frac{2T - T'}{10} = K \left( \frac{2T + T'}{2} - T \right)$ $\frac{2T - T'}{10} = K \left( \frac{2T + T' - 2T}{2} \right)$ $\frac{2T - T'}{10} = K \left( \frac{T'}{2} \right)$ ... (ii)

Dividing equation (i) by (ii), we get: $\frac{\frac{T}{10}}{\frac{2T - T'}{10}} = \frac{K \left( \frac{3T}{2} \right)}{K \left( \frac{T'}{2} \right)}$ $\frac{T}{2T - T'} = \frac{3T}{T'}$ $T' = 3(2T - T')$ $T' = 6T - 3T'$ $4T' = 6T$ $T' = \frac{6T}{4} = \frac{3}{2}T$

Therefore, the temperature of the body at the end of next $10\text{ minutes}$ will be $\frac{3}{2}T$.