How to determine the calorimeter constant from its heat capacity
To give a precise answer, the following assumptions are necessary and must be clear:


bomb calorimeter works at constant volume ($V=const$);
both water and calorimeter itself are at thermodynamic equilibrium before the experiment and during the measurement, in particular their temperatures $T_w$ and $T_c$ are equal before the experiment and during the measurement;
the system is compound by calorimeter itself plus water;
the system is an isolated one;
pressure is 1 bar.




Initially the system is at temperature $T_1$. Let's imagine that an object at $T_o&gt;T_1$ is put inside the chamber of the calorimeter. Temperature of the system increases and, once reached thermodynamic equilibrium, it stops at a precise value $T_2$.

Since $V=const$, heat transferred from object to system is:
\begin{equation}
Q_V=\Delta U=\Delta U_{calorimeter}+\Delta U_{water}=(mc_V\Delta T)_c+(mc_V\Delta T)_w
\end{equation}
where $\Delta T_c=\Delta T_w=T_2-T_1$.

We know that heat capacity at constant volume is defined as:
\begin{equation}
C_V=\left(\frac{\partial U}{\partial T}\right)_V\approx \left(\frac{\Delta U}{\Delta T}\right)_V
\end{equation}
So, reshaping the first equation, we obtain:
\begin{equation}
C_V=\frac{\Delta U}{\Delta T}=(mc_V)_c+(mc_V)_w=(C_V)_c+(\rho Vc_V)_w
\end{equation}
Adding the following data:


$\rho_w=1000\;kg/m^3$;
$(c_V(300\;K,1\;bar))_w\approx 4.134\;J/(kg\;K)$ (source: Perry's Chemical Engineers' Handbook)


and carrying out the conversion: $V=600\;mL=6\times10^{-4}\;m^3$, we obtain finally:
\begin{equation}
C_V=787\;J/K=0.787\;kJ/K
\end{equation}
So right answer is A.