Diborane burns completely in oxygen: B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g)…

Step 1. Moles of diborane. $1.12 / 22.4 = 0.0500\ \text{mol}$.

Step 2. Oxygen. The equation needs $3$ mol of O$_2$ for each mole of diborane, so $0.150\ \text{mol}$. The volume is $0.150 \times 22.4 = 3.36\ \text{L}$.

Step 3. Boron(III) oxide. One mole forms for each mole of diborane: $0.0500 \times 69.62 = 3.48\ \text{g}$.

Step 4. The enthalpy of the reaction. $\Delta H = [-1273.5 + 3(-241.8)] - [36.4] = -2035.3\ \text{kJ}$ for one mole of diborane.

Step 5. The heat. $0.0500 \times 2035.3 = 101.8\ \text{kJ}$ is released.

Source. The enthalpies of formation are standard values from the NIST Chemistry WebBook and the CRC Handbook. NCERT prints $-1976\ \text{kJ/mol}$ for the combustion of diborane. The answer here follows the standard formation enthalpies given in the statement.