Solution:

Given:
- Mass of water \( m_w = 0.2 \, \text{kg} \)
- Initial temperature of water \( T_w = 80^\circ \text{C} \)
- Final temperature \( T_f = 60^\circ \text{C} \)
- Specific heat of water \( c_w = 4200 \, \text{J/kg}^\circ \text{C} \)
- Mass of solid \( m_s = 0.2 \, \text{kg} \)
- Initial temperature of solid \( T_s = 20^\circ \text{C} \)
- Specific heat of solid \( c_s \) (to be found)

Heat lost by water:
\[
Q_{\text{lost (water)}} = m_w c_w (T_w - T_f) = 0.2 \times 4200 \times (80 - 60) = 16800 \, \text{J}
\]

Heat gained by solid:
\[
Q_{\text{gained (solid)}} = m_s c_s (T_f - T_s) = 0.2 \times c_s \times (60 - 20) = 8 c_s
\]

From the heat exchange equation:
\[
16800 = 8 c_s
\]
\[
c_s = \frac{16800}{8} = 2100 \, \text{J/kg}^\circ \text{C}
\]

Since the specific heat of water is \( 4200 \, \text{J/kg}^\circ \text{C} \), we see that the specific heat of the solid is:

\[
c_s = \frac{4200}{2} = 2100 \, \text{J/kg}^\circ \text{C}
\]

Thus, the specific heat of the solid is indeed half of the specific heat of water.