Here is the balanced equation:
$C_6H_12O_6 + 6O_2 rarr 6 CO_2 + 6 H_2O$
Since the molar mass of glucose is 180 g, the 45 g given mass in this problem is
$"no. of moles" = (45g)/(180g) = 0.25$ mole
The equation tells us there are 6 moles of $CO_2$ produced for each 1 mole of glucose that reacts. There fore, we will obtain
$6xx0.25 = 1.50 "moles of " C_6H_12O_6$
Finally, converting to mass
$1.50 "moles" xx 44 g/"mole" =66 g$