Need help, chemistry!

Calculate the enthalpy of the following reaction: 4 b (s) + 3 o2 (g) → 2 b2o3 (s) given the following pertinent information: (a) b2o3 (s) + 3 h2o (g) → 3 o2 (g) + b2h6 (g), δhoa = +2035 kj (b) 2 b (s) + 3 h2 (g) → b2h6 (g), δhob = +36 kj (c) h2 (g) + latex: \frac{1}{2} 1 2 o2 (g) → h2o (l), δhoc = −285 kj (d) h2o (l) → h2o (g), δhod = +44 kj

I’m an electrically neutral atomic any element, there are equal numbers of

37.0 °c = °f (human body normal temperature)