The answer is #75.6 grams#.
We are dealing with boiling-point elevation, which states that a solution has a higher boiling point than a pure solvent; that is, a solvent's boiling point (in this case, water's) will increase with the adition of, in this case, ethylene glycol.
The ecuation that models boiling-point elevation is
#DeltaT_(boil i ng) = k_b * m#, where
#DeltaT# - boiling point elevation (#T_(f i nal) - T_(i nitial)#);
#k_b# - ebullioscopic constant;
#m# - molality (NOT mass);
Water's ebullioscopic constant is usually given, at its normal boiling point being #k_b# = #0.512 (degrees C * kg)/(mo l e)#.
We know that #DeltaT_b# = #102.5 - 100# = #2.5^@ C#, therefore we can determine the solution's molality
#m = (DeltaT_b)/k_b = (2.5^@ C)/((0.512^@C * kg)/(mol e)) = 4.88 (mol es)/(kg)#. However, molality is defined as moles of solute devided by mass of solvent (in kg), which means
#m = n_(solute)/(mass_(solvent)) -> n_(solute) = m * mass_(solute) = 4.88 (mol es)/(kg) * 250 * 10^(-3) kg = 1.22 mol es#
Knowing that ethylene glycol's molar mass is #62 g/(mol e)#, we can get its mass by
#mass = n * molarmass = 1.22 mol es * 62 g/(mol e) = 75.6 grams#.
