To find the sum of the series given in the image, we need to analyze the pattern and simplify the terms.

The series is:
\[ E = \frac{1}{13} + \frac{3}{13 \cdot 23} + \frac{3}{23 \cdot 33} + \cdots + \frac{3}{1993 \cdot 2003} \]

This series can be written in a general form:
\[ E = \sum_{n=0}^{199} \frac{3}{(10n+13)(10n+23)} \]

We can use partial fraction decomposition to simplify each term:
\[ \frac{3}{(10n+13)(10n+23)} = \frac{A}{10n+13} + \frac{B}{10n+23} \]

Solving for \(A\) and \(B\):
\[ 3 = A(10n+23) + B(10n+13) \]

Setting up the system of equations:
\[ A(10n+23) + B(10n+13) = 3 \]

By comparing coefficients, we get:
\[ A + B = 0 \]
\[ 23A + 13B = 3 \]

Solving these equations:
\[ B = -A \]
\[ 23A - 13A = 3 \]
\[ 10A = 3 \]
\[ A = \frac{3}{10} \]
\[ B = -\frac{3}{10} \]

So, the partial fractions are:
\[ \frac{3}{(10n+13)(10n+23)} = \frac{\frac{3}{10}}{10n+13} - \frac{\frac{3}{10}}{10n+23} \]
\[ = \frac{3}{10} \left( \frac{1}{10n+13} - \frac{1}{10n+23} \right) \]

Thus, the series becomes:
\[ E = \frac{3}{10} \left( \sum_{n=0}^{199} \left( \frac{1}{10n+13} - \frac{1}{10n+23} \right) \right) \]

This is a telescoping series, where most terms cancel out:
\[ E = \frac{3}{10} \left( \left( \frac{1}{13} - \frac{1}{23} \right) + \left( \frac{1}{23} - \frac{1}{33} \right) + \cdots + \left( \frac{1}{1993} - \frac{1}{2003} \right) \right) \]

After cancellation, we are left with:
\[ E = \frac{3}{10} \left( \frac{1}{13} - \frac{1}{2003} \right) \]

Simplifying:
\[ E = \frac{3}{10} \left( \frac{2003 - 13}{13 \cdot 2003} \right) \]
\[ E = \frac{3}{10} \left( \frac{1990}{13 \cdot 2003} \right) \]
\[ E = \frac{3 \cdot 1990}{10 \cdot 13 \cdot 2003} \]
\[ E = \frac{5970}{26039} \]

Therefore, the sum of the series is:
\[ E = \frac{5970}{26039} \]