Refer to the circuit diagram given below. The heat generated across...

Taking the right-hand wire as the reference node at $0\text{ V}$, let the left junction (where the $3\,\Omega$, $4\,\Omega$ and $6\,\Omega$ resistors meet) be Node A at potential $V_A$.

Bottom branch: The positive terminal of the $2\text{ V}$ battery is at the reference, so its negative terminal (and hence the left end of the $3\,\Omega$ resistor) is at $-2\text{ V}$.
Top branch: The positive terminal of the $3\text{ V}$ battery is at the reference, so its negative terminal (and hence the node between the $6\,\Omega$ resistor and the $3\text{ V}$ battery) is at $-3\text{ V}$.

Applying KCL at Node A (sum of currents leaving the node is zero):
$\dfrac{V_A - 0}{4} + \dfrac{V_A - (-2)}{3} + \dfrac{V_A - (-3)}{6} = 0$

$\dfrac{V_A}{4} + \dfrac{V_A + 2}{3} + \dfrac{V_A + 3}{6} = 0$

Multiplying throughout by $12$:
$3V_A + 4(V_A + 2) + 2(V_A + 3) = 0$

$3V_A + 4V_A + 8 + 2V_A + 6 = 0$

$9V_A + 14 = 0 \Rightarrow V_A = -\dfrac{14}{9}\text{ V}$

Current through the $6\,\Omega$ resistor (from Node A towards the $-3\text{ V}$ point):
$I_6 = \dfrac{V_A - (-3)}{6} = \dfrac{-\dfrac{14}{9} + 3}{6} = \dfrac{\dfrac{13}{9}}{6} = \dfrac{13}{54}\text{ A}$

Heat generated in the $6\,\Omega$ resistor in $t = 100\text{ s}$:
$H = I_6^2 \cdot R \cdot t = \left(\dfrac{13}{54}\right)^2 \cdot 6 \cdot 100$

$H = \dfrac{169}{2916} \cdot 600 = \dfrac{169 \cdot 100}{486} = \dfrac{16900}{486}\text{ J}$

Given $H = \dfrac{\alpha}{100}\text{ J}$:
$\dfrac{\alpha}{100} = \dfrac{16900}{486}$

$\alpha = \dfrac{16900 \cdot 100}{486} = \dfrac{1690000}{486} \approx 3477.37$

Rounding to the nearest integer:
$\alpha = 3477$

Hence, the answer is $3477$.