Calculating the viscosity of water using particles
The basic idea here is that the spherical particles, each of radius a, are being bounced around via Brownian motion. So the two relevant equations are
$$r(t)^2 = 2Dt \tag{1}$$
$$D = \frac{kT}{6πηa} \tag{2}$$
Note that the first equation uses r(t) rather than x, to avoid needless confusion with the x and y cartesian coordinates of the spherical particles.
Now consider particle 1. At t = 0 s, it is at the defined location 0,0, where all distances are in $\mu m$. At four subsequent times, the x and y coordinates are as given in the OP's table. The following figure shows particle 1's location at t = 0s and at t = 30 s.

At any given time,
$$r_1(t)^2 = x_1(t)^2 + y_1(t)^2 \tag{3}$$
To estimate D, an ordinary least squares fit is performed on the tabulated data for particle 1. The result is shown in the next figure:

Note that the origin is used: at t = 0 s, then, by definition, $x_1(0) = 0 \space µm$ and $y_1(0) = 0 \space µm$. So $D = 1.1486 \times 10^{-12} m^2 s^{-1}$. Rounding off to two significant figures will be performed at the end.
Solving equation (2) for $\eta $ yields:
$$\eta = \frac{kT}{6πDa} \tag{4}$$
where T = 298.15 K, $a = 2.12 \times 10^{-7} m$ and $k = 1.38064852 \times 10^{-23} J K^{-1} $. The result for $\eta $ is
$$\eta = 8.97 \times 10^{-4} Nt \space m^{-2} \space s \tag{5}$$
where the fact that $1 J = 1 Nt \times 1 m$ has been used. N.B. 1 Joule equals 1 Newton times 1 meter.
Other units for $\eta $: Since 1 Pascal of pressure is 1 Newton of force per square meter, i.e., $1 Pa = 1 Nt / m^2$, and since 1 centipoise (symbol: cP) is 1 milliPascal times 1 s, i.e., $1 cP = 1 mPa \space s$, then
$$\eta = 8.97 \times 10^{-4} Pa \space s = 0.897 \space mPa \space s = 0.897 cP\tag{6}$$
Finally, rounding to two significant figures, $\eta = $0.90 cP. The viscosity of water at 25 °C is reported to be 0.91 cP.