You need to perform the following hypothesis test:
\begin{align} H_0 &: \mu_1 = \mu_2 \\ H_a &: \mu_1 \neq \mu_2 \end{align}
where $H_0$ and $H_a$ are the null and alternate hypothesis respectively. $\mu_1$ and $\mu_2$ are the means of the two samples. You will have to compute the following test statistic:
\begin{align} t &= \frac{\hat{\mu}_d}{\sqrt{\hat{s}_d^2/n}} \end{align}
where $n$ is the number of samples ($63$ in your case), $\hat{\mu}_d$ and $\hat{s}_d^2$ are the sample mean and sample variance of the difference between the paired readings.
The above test statistic has $n-1$ degrees of freedom. So assuming you want to check for significance of the difference between the two means at $5\%$ level, you will have to:
- Compute the statistic using your data.
- Compare the resulting number to $2$.
- If your statistic is greater than $2$, the null hypothesis is rejected and there exists a significant difference between the two means at the $5\%$ level. Else, you cannot reject the null hypothesis.
In (2), I have given you the critical value by reading off the t-distribution table here: http://www.math.unb.ca/~knight/utility/t-table.htm. I suggest you to review basic statistical hypothesis testing from a standard text book (e.g. Hogg) in case you want a more detailed understanding.