Looks easy enough. A PF= 1.0 implies a purely resistive load. If the load is the series inductor and resistor, then:
Zl= 7.0 + j26.39 R
Since it's a parallel connection, you need the equivalent admittance of the load:
Yl= 1/Zl
Yl= 1/(7 + j26.39)= 9.39E-3 - j35.4E-3 A/V
Knowing you have a Bl= -j35.4E-3, you need a Bc= j35.4E-3:
Bc= wC:
C= Bc/w= 35.4E-3/(2pi * 60)= 93.9uF