A term in the expansion of $(1+\sqrt{2}+\sqrt[3]{5}{)}^6$ is

of the form

$\frac{6!}{r!s!(6-r-s)!}\left(1{)}^{6-r-s}\right(\sqrt{2}{)}^r(\sqrt[3]{5}{)}^s$

$=\frac{6!}{r!s!(6-r-s)!}\left(2^{r/2}\right)\left(5^{s/3}\right)$

This term will be rational if $2^{r/2}$ and $5^{s/3}$ are both rational numbers. This is possible if and only if  $r$ is $a$ multiple of $2$ and s is a multiple of $3.$ Possible values of r are $0, 2, 4$ and 6 whereas the possible values of $s$ are  $0, 3 6$ and  Also note 

that $0\le r+s\le 6$

For $r=0,s$ can take value $0, 3$ or $6$

For$r = 2, s$ can take value 0 or $3$

For$r = 4, s$ can take value $0$

For $r = 6, s$ can take value $0.$

Thus, the number of rational terms is  $3+2+1+1=7$