A 6 V ideal battery, a 1 resistor, and a 2.0 resistor are connected...

Based on the problem description, the $6\text{ V}$ battery, the $1\text{ }\Omega$ resistor, and the $2.0\text{ }\Omega$ resistor are connected in series.

The equivalent resistance of the circuit is:
$R_{eq} = 1\text{ }\Omega + 2.0\text{ }\Omega = 3\text{ }\Omega$

The current flowing through the circuit is:
$I = \dfrac{V}{R_{eq}} = \dfrac{6}{3} = 2\text{ A}$

The power dissipated by the $2.0\text{ }\Omega$ resistor (which is inside the calorimeter) is:
$P = I^2 R = (2)^2 \times 2.0 = 8\text{ W}$

The heat generated in the $2.0\text{ }\Omega$ resistor over a time $t = 15\text{ minutes} = 15 \times 60\text{ s} = 900\text{ s}$ is:
$H = P \times t = 8 \times 900 = 7200\text{ J}$

The rise in temperature $\Delta T$ of the calorimeter and water is given by:
$H = C \Delta T$
where $C = 2000\text{ J K}^{-1}$ is the heat capacity.

$\Delta T = \dfrac{H}{C} = \dfrac{7200}{2000} = 3.6\text{ K}$

A temperature rise of $3.6\text{ K}$ is equivalent to a rise of $3.6^\circ\text{C}$.

Answer: $3.6^\circ\text{C}$