Add HBR/ROOR/∆ to form Anti-Markonikov product.

Then, add t-BuOK/∆ to form anti-Satzeff product. t-BuOK/∆ is a bulky base so it can't remove H from the 3° Carbon, instead it removes hydrogen from the 1°Carbon and forms Anti-Satzeff product.

Finally, we have to do Hydro-Boration Reaction in presence of H2O2/NaOH.

Adding the respective numbers 6+10+4+9= 29.