Solution:

To find the water equivalent of the calorimeter, we can use the principle of heat exchange:

\[
\text{Heat lost by hotter water} = \text{Heat gained by colder water} + \text{Heat gained by the calorimeter}
\]

Let \( m_1 = 0.2 \, \text{kg} \), \( T_1 = 30^\circ \text{C} \) (colder water), \( m_2 = 0.1 \, \text{kg} \), \( T_2 = 60^\circ \text{C} \) (hotter water), and the final temperature \( T_f = 35^\circ \text{C} \). Let \( W_c \) be the water equivalent of the calorimeter, and \( c = 4200 \, \text{J/kg}^\circ \text{C} \) be the specific heat capacity of water.

Heat lost by hotter water:
\[
Q_{\text{lost}} = m_2 c (T_2 - T_f) = 0.1 \times 4200 \times (60 - 35) = 0.1 \times 4200 \times 25 = 10500 \, \text{J}
\]

Heat gained by colder water:
\[
Q_{\text{gained (water)}} = m_1 c (T_f - T_1) = 0.2 \times 4200 \times (35 - 30) = 0.2 \times 4200 \times 5 = 4200 \, \text{J}
\]

Heat gained by calorimeter:
\[
Q_{\text{gained (calorimeter)}} = W_c \times c \times (T_f - T_1) = W_c \times 4200 \times 5
\]

Using heat balance equation:
\[
Q_{\text{lost}} = Q_{\text{gained (water)}} + Q_{\text{gained (calorimeter)}}
\]
\[
10500 = 4200 + W_c \times 4200 \times 5
\]
\[
10500 - 4200 = 21000 W_c
\]
\[
6300 = 21000 W_c
\]
\[
W_c = \frac{6300}{21000} = 0.3 \, \text{kg}
\]

Since the water equivalent is in terms of mass and \( c = 4200 \, \text{J/kg}^\circ \text{C} \), the water equivalent in \( \text{J/K} \) is:
\[
W_c \times c = 0.3 \times 4200 = 1260 \, \text{J/K}
\]

So, the water equivalent of the calorimeter is 1260 J/K.