On complete combustion, 0 . 492g of an organic compound gave 0 ....

$44$ grams of ${CO}_{2}$ contain $12$ grams of carbon.

Hence, one gram of carbon dioxide contain $\dfrac{12}{44}$ gram of carbon.

$0 . 792$ gram of carbon dioxide contain $\dfrac{12}{44} \times 0 . 792$ gram of carbon

Percentage of carbon = $\dfrac{\text{Mass of carbon}}{\text{Mass of organic compound}} \times 100$

$= \dfrac{12}{44} \times \dfrac{0 . 792}{0 . 492} \times 100$

$= 43 . 90$

So, answer is $44$.