$foo(1037,2)$: Current value of $n=1037$, $d=2$
inside the function $foo(1037,2)$ $x$ is a variable initialized with $0$
Next there is a while loop which is run till $(n\geq 1)$,loop get terminated when $n$ value is smaller then $1$.
Working of while loop is as follows:
$1^{st}$ iteration: $(1037\geq1)=T$,
$x$ is incremented by $1$ that is $x=1$.
Current value of $n=518$
$2^{nd}$ iteration: $(518 \geq1)=T$,
Current value of $x=2$,
Current value of $n=259$
$3^{rd}$ iteration: $(259 \geq1)=T$,
Current value of $x=3$,
Current value of $n=129$
$4^{th}$ iteration: $(129 \geq1)=T$,
Current value of $x=4$,
Current value of $n=64$
$5^{th}$ iteration: $(64 \geq1)=T$,
Current value of $x=5$,
Current value of $n=32$
$5^{th}$ iteration: $(64 \geq1)=T$,
Current value of $x=5$,
Current value of $n=32$
$6^{th}$ iteration: $(32 \geq1)=T$,
Current value of $x=6$,
Current value of $n=16$
$7^{th}$ iteration: $(16 \geq1)=T$,
Current value of $x=7$,
Current value of $n=8$
$8^{th}$ iteration: $(8 \geq1)=T$,
Current value of $x=8$,
Current value of $n=4$
$9^{th}$ iteration: $(4 \geq1)=T$,
Current value of $x=9$,
Current value of $n=2$
$10^{th}$ iteration: $(2 \geq1)=T$,
Current value of $x=10$,
Current value of $n=1$
$11^{th}$ iteration: $(1 \geq1)=T$,
Current value of $x=11$,
Current value of $n=0$
$12^{th}$ iteration: $(0 \geq1)=F$,
exit from while loop. So current value of $x=11$
So $x=11$ is correct answer.