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Solution

$n=pq=3\times11=33$

$\phi(n)=(p-1)(q-1)=2\times10=20$

Now,

$d\times e\equiv1\pmod{\phi(n)}$

$7e\equiv1\pmod{20}$

$e=3$, because $7\times3=21\equiv1\pmod{20}$

Now encrypt $M=19$:

$C=M^e\mod n$

$C=19^3\mod33$

$19^2=361\equiv31\pmod{33}$

$19^3=31\times19=589\equiv28\pmod{33}$

So, $e=3$ and encrypted value $=28$.