The greatest number which divides $990$ and $1330$, leaving remainder $6$ and $10$ respectively is:

The required greatest number is the $\text{HCF}$ of $(990-6)$ and $(1330-10)$.
$$990-6=984$$

$$1330-10=1320$$

Let's find the $\text{HCF}(984,1320)$ using the division method:

$$1320=984\times 1+336$$

$$984=336\times 2+312$$

$$336=312\times 1+24$$

$$312=24\times 13+0$$

Since the remainder becomes $0$ when divided by $24$, the $\text{HCF}$ is $24$.