The formula for the standard deviation of a set of numbers is given by:

\[

\sigma = \sqrt{\frac{\sum_{i=1}^n (x_i - \mu)^2}{n}}

\]

Where:

- \(\sigma\) is the standard deviation,

- \(x_i\) are the individual values,

- \(\mu\) is the mean of the values,

- \(n\) is the number of values.

In this case, we have the values \(2t\), \(3t\), \(4t\), \(5t\), and \(6t\), and we are given that the standard deviation is \(\sqrt{2}\).

We need to find the value of \(t\) for which this holds true. The first step is to calculate the mean of the values:

\[

\mu = \frac{(2t + 3t + 4t + 5t + 6t)}{5} = \frac{20t}{5} = 4t

\]

Now, we can plug in the values to the standard deviation formula:

\[

\sqrt{2} = \sqrt{\frac{(2t - 4t)^2 + (3t - 4t)^2 + (4t - 4t)^2 + (5t - 4t)^2 + (6t - 4t)^2}{5}}

\]

Simplify the numerator:

\[

\sqrt{2} = \sqrt{\frac{4t^2 + t^2 + 0 + t^2 + 4t^2}{5}}

\]

\[

\sqrt{2} = \sqrt{\frac{10t^2}{5}}

\]

\[

\sqrt{2} = \sqrt{2t^2}

\]

Now, square both sides to solve for \(t\):

\[

2 = 2t^2

\]

Divide by 2:

\[

t^2 = 1

\]

Take the square root:

\(

t = + 1

\)