Answer
Please see the work below.
Work Step by Step
We know that
$Q_{water}=Q_{boiling}$
$\implies mc\Delta T=(\frac{1}{10})mL_v$
$mc(100-T_i)=(\frac{1}{10}mL_v)$
$c(100-T_i)=(\frac{1}{10}L_v)$
We plug in the known values to obtain:
$(4184)(100-T_i)=(\frac{1}{10})(2257000)$
This simplifies to:
$T_i=46^{\circ}C$
