What you need to calculate here is the 95% confidence interval ($1-\alpha=0.95$) for the mean difference. For large samples, it is given by:
$$\left[\bar{X}_{1}-\bar{X}_{2}-z_{1-\frac{\alpha}{2}}\sqrt{\frac{\sigma_{1}^2}{n_{1}}+\frac{\sigma_{2}^2}{n_{2}}},\bar{X}_{1}-\bar{X}_{2}+z_{1-\frac{\alpha}{2}}\sqrt{\frac{\sigma_{1}^2}{n_{1}}+\frac{\sigma_{2}^2}{n_{2}}}\right].$$
Intuitively, it is the interval in which the true value of the mean of the difference of the fill volumes is located with 95% likelihood.
More precisely, you are interested in the upper bound, i.e. $\bar{X}_{1}-\bar{X}_{2}+z_{1-\frac{\alpha}{2}}\sqrt{\frac{\sigma_{1}^2}{n_{1}}+\frac{\sigma_{2}^2}{n_{2}}}$.
Now, you were wondering what $z_{1-\frac{\alpha}{2}}$. It such value of $z$, which satisifies $\Phi(z)=1-\frac{\alpha}{2}$ where $\Phi(\cdot)$ is the cumulative distribution function of the standard normal distribution. The value given by our colleague, $1.96$, was the correct value, retrieved from the tables for standard normal distribution.
However, as your values for both $n$ are rather low, it seems reasonable to use the t-distribution instead of normal distribution. You upper bound changes thus to:
$$\bar{X}_{1}-\bar{X}_{2}+t\left(\frac{\alpha}{2},\nu\right)\sqrt{\frac{\sigma_{1}^2}{n_{1}}+\frac{\sigma_{2}^2}{n_{2}}}$$
where $\alpha/2$ is the upper tail probability ($1-\alpha=0.95$ is the confidence level) and $\nu=n_1+n_2-2$ are the degrees of freedom. You may find in the tables of t-distribution that $t(0.025,20)=2.086$ (though you may find other notations, such as $t_{0.975}$ - it usually follows from the table which notation is used). Plug in the rest of your numbers, and you should obtain the correct result.
