How to calculate the heat capacity of a calorimeter based on a neutralisation reaction?
First, you have to find out the concentrations so you can calculate the amount of heat released. Since the density is given for a 1 M solution, I will assume this concentration.

Next, the mass of our solution:
$$ m = \rho \times V = 103.7~\mathrm{g}$$

Now we have to calculate the amount of energy that gets released through the reaction, but viewed from the environment and not the reaction (that's the reason for the negative sign before $q$):
$$ -q = \Delta_\text{r}H \times n = -58.3 \times 0.1 ~\mathrm{kJ\, mol^{-1}\, mol} = -5.83~\mathrm{kJ}$$

The total heat capacity of your system is given as
$$ c_\text{tot} = c_\text{s, sol}\times m + c_\text{cal}$$
where $c_\text{s, sol}$ is the specific heat capacity of your solution and $c_\text{cal}$ is the heat capacity of the calorimeter.

Now we know all the quantities needed for the following equation:
$$ q = c_\text{tot}\Delta T$$

We expand:
$$ q =( c_\text{s, sol}\times m + c_\text{cal}) \times \Delta T$$

Solve for $c_\text{cal}$:
$$ c_\text{cal} = \frac{q-c_\text{s, sol}m\Delta T}{\Delta T} = \frac{5.83~\mathrm{kJ} - 3.91~\mathrm{J\, g^{-1}\, K^{-1}} \times 103.7~\mathrm{g} \times 11.5~\mathrm{K}}{11.5~\mathrm{K}} = 101~\mathrm{J\, K^{-1}}$$

And now you know the heat capacity of the calorimeter.