Given: 

M\(_r\) of MgSO\(_4\).7H\(_2\)O = 246

M\(_r\) of Oxygen = 16

Percentage by mass of Oxygen =  \(\frac{Total mass of Oxygen}{Total mass of MgSO_4.7H_2O}\times 100\)

% by mass of Oxygen = \(\frac{11moles of oxygen}{MgSO_4.7H_2O}\times 100\) (4moles of oxygen +  7 moles of oxygen from 7H\(_2\)O )

  = \(\frac{11 X 16}{246}\times 100\)

  = \(\frac{176}{246}\times 100\)

  = 71.544%

  = 71.5%