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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

$1.50 "moles" xx 44 g/"mole" =66 g$

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