Answer to the Question
In section 3 of his paper Elliptic Curve Cryptosystems, Koblitz describes three methods to encode plaintext messages as elliptic curve points. Method (2) is the one usually described as Koblitz's method and assumes that we are working over a prime field and that plaintext is represented as a non-negative integer $m$ less than $p/1000-1$. To encode one tests in turn putative $x$ co-ordinates $x=1000m, 1000m+1,\ldots,1000m+999$ until we find one where $x^3+ax+b$ is a square modulo $p$ and then compute the corresponding $y$. Decoding is achieved by deleting the last 3 digits of the $x$-coordinate of an elliptic curve point.
The condition $0\le m<p/1000-1$ sets a limit on the size of the message and long messages will have to be encoded in multiple points. If we bound the size of integers as $B<p/1000-1$, writing the message in base-$B$ allows us to encode a message cooresponding to the integer $L$ in $\lfloor\log L/\log B\rfloor+1$ points.
Additional information beyond the scope of the question
The value 1000 is somewhat arbitrary, but should be chosen to minimise the chance of failure to encode. As each of our candidate $x$ values will produce a square value of $x^3+ax+b$ with heuristic probability close to 1/2, we might expect that this method to fail with probability $\approx 2^{-1000}$ which is highly improbable. Replacing 1000 with a smaller number $k$ (and changing the decoding procedure accordingly) will give an increased plaintext space by a factor $\approx 1000/k$ failure rate of $2^-k$. One should consider the effect of adversarially chosen messages before selecting $k$ too small.
One should also note that the decoding process works whether the correct encoding process was followed or not. This means that multiple points (roughly 500) on the curve decode to the same plaintext. This could provide a covert channel in a malicious implementation.
