By definition, $"1 cal"$ is the approximate amount of energy needed to raise the temperature of $"1 g"$ of water by $1^@ "C"$ at $"1 atm"$.
In thermochemistry, this is known to be equal to $"4.184 J"$ at $25^@ "C"$. Therefore...
$5.25 cancel"cal" xx "4.184 J"/cancel"cal" = "21.97 J"$
Or to three sig figs, $~~$ $ulcolor(blue)"22.0 J"$.